Given a particle in one dimension subject to a constant force F with Hamilton function H = p²/2m − Fx, find x(t) by using the Lie series (8.147).
u\left(t\right) =u\left(0\right)+t\left\{u,H\right\}_{0} +\frac{t^{2}}{2!}\left\{\left\{u,H\right\},H\right\} _{0}+ \frac{t^{3}}{3!}\left\{\left\{\left\{u,H\right\},H\right\},H\right\} _{0}+· · · . (8.147)
We have
{x,H} = \frac{1}{2m}\left\{x, p^{2}\right\}=\frac{1}{2m}2p\left\{x, p\right\}= \frac{p}{m}, (8.149)
\left\{\left\{x, H\right\},H\right\}= \frac{1}{m}\left\{p, H\right\}= \frac{-F}{m}\left\{p,x\right\}=\frac{F}{m}. (8.150)
Since F = constant, all the next Poisson brackets vanish and the Lie series (8.147) with u = x gives
x(t) =x_{0}+t\left\{x,H\right\}_{0}+\frac{t^{2}}{2!}\left\{\left\{x,H\right\},H\right\}_{0}=x_{0}+ \frac{p_{0}}{m}t+ \frac{F}{2m}t^{2}, (8.151)
which is the well-known elementary solution.